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LECTURE 26: FEEDBACK CONTROL. Objectives: Typical Feedback System Feedback Example Feedback as Compensation Proportional Feedback Applications

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LECTURE 26: FEEDBACK CONTROL


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slide1

LECTURE 26: FEEDBACK CONTROL

  • Objectives:Typical Feedback SystemFeedback ExampleFeedback as CompensationProportional FeedbackApplications
  • Resources:MIT 6.003: Lecture 20MIT 6.003: Lecture 21Wiki: Control SystemsBrit: Feedback ControlJC: Crash CourseWiki: Root LocusWiki: Inverted Pendulum CJC: Inverted Pendulum

Audio:

URL:

slide2

A Typical Feedback System

Feed Forward

Feedback

  • Why use feedback?
  • Reducing Nonlinearities
  • Reducing Sensitivity to Uncertainties and Variability
  • Stabilizing Unstable Systems
  • Reducing Effects of Disturbances
  • Tracking
  • Shaping System Response Characteristics (bandwidth/speed)
slide3

Motivating Example

  • Open loop system: aim and shoot.
  • What happens if you miss?
  • Can you automate the correctionprocess?
  • Closed-loop system: automatically adjusts until the proper coordinates are achieved.
  • Issues: speed of adjustment, inertia, momentum, stability, …
slide4

System Function For A Closed-Loop System

  • The transfer function of thissystem can be derived usingprinciples we learned inChapter 6:
  • Black’s Formula: Closed-loop transfer function is given by:
  • Forward Gain: total gain of the forward path from the inputto the output, where the gain of a summer is 1.
  • Loop Gain: total gain along the closed loop shared by all systems.

Loop

slide5

The Use Of Feedback As Compensation

  • Assume the open loopgain is very large(e.g., op amp):

 Independent of P(s)

  • The closed-loop gain depends only on the passive components (R1 and R2) and is independent of the open-loop gain of the op amp.
slide6

Stabilization of an Unstable System

  • If P(s) is unstable, can westabilize the system byinserting controllers?
  • Design C(s) and G(s) so thatthe poles of Q(s) are in the LHP:
  • Example: Proportional Feedback (C(s) = K)
  • The overall system gain is:
  • The transfer function is stable for K > 2.
  • Hence, we can adjust K until the system is stable.
slide7

Second-Order Unstable System

  • Try proportional feedback:
  • One of the poles is at
  • Unstable for all values of K.
  • Try damping, a term proportional to :
  • This system is stable as long as:
  • K2 > 0: sufficient damping force
  • K1 > 4: sufficient gain
  • Using damping and feedback, we have stabilized a second-order unstable system.
slide8

The Concept of a Root Locus

  • Recall our simple control systemwith transfer function:
  • The controllers C(s) and G(s) can bedesigned to stabilize the system, but that could involve a multidimensional optimization. Instead, we would like a simpler, more intuitive approach to understand the behavior of this system.
  • Recall the stability of the system depends on the poles of 1 + C(s)G(s)P(s).
  • A root locus, in its most general form, is simply a plot of how the poles of our transfer function vary as the parameters of C(s) and G(s) are varied.
  • The classic root locus problem involves a simplified system:

Closed-loop poles are the same.

slide9

Example: First-Order System

  • Consider a simple first-order system:
  • The pole is at s0 = -(2+K). Vary Kfrom 0 to :
  • Observation: improper adjustment of the gain can cause the overall system to become unstable.

Becomes less stable

Becomes more stable

slide10

Example: Second-Order System With Proportional Control

  • Using Black’s Formula:
  • How does the step responsevary as a function of the gain, K?
  • Note that as K increases, thesystem goes from too little gainto too much gain.
slide11

How Do The Poles Move?

Desired Response

  • Can we generalize this analysis to systems of arbitrary complexity?
  • Fortunately, MATLAB has support for generation of the root locus:
  • num = [1];
  • den = [1 101 101]; (assuming K = 1)
  • P = tf(num, den);
  • rlocus(P);
slide14

Summary

  • Introduced the concept of system control using feedback.
  • Demonstrated how we can stabilize first-order systems using simple proportional feedback, and second-order systems using damping (derivative proportional feedback).
  • Why did we not simply cancel the poles?
  • In real systems we never know the exact locations of the poles. Slight errors in predicting these values can be fatal.
  • Disturbances between the two systems can cause instability.
  • There are many ways we can use feedback to control systems including feedback that adapts over time to changes in the system or environment.
  • Discussed an application of feedback control involving stabilization of an inverted pendulum.
slide15

More General Case

  • Assume no pole/zero cancellation in G(s)H(s):
  • Closed-loop poles are the roots of:
  • It is much easier to plot the root locus for high-order polynomials because we can usually determine critical points of the plot from limiting cases(e.g., K= 0, ), and then connect the critical points using some simple rules.
  • The root locus is defined as traces of s for unity gain:
  • Some general rules:
  • At K= 0, G(s0)H(s0) =   s0are the poles of G(s)H(s).
  • At K= , G(s0)H(s0) = 0 s0are the zeroes of G(s)H(s).
  • Rule #1: start at a pole at K= 0 and end at a zero at K= .
  • Rule #2: (K  0) number of zeroes and poles to the right of the locus point must be odd.
slide16

Inverted Pendulum

  • Pendulum which has its mass above its pivot point.
  • It is often implemented with the pivot point mounted on a cart that can move horizontally.
  • A normal pendulum is stable when hanging downwards, an inverted pendulum is inherently unstable.
  • Must be actively balanced in order to remain upright, either by applying a torque at the pivot point or by moving the pivot point horizontally (Wiki).
slide17

Feedback System – Use Proportional Derivative Control

  • Equations describing the physics:
  • The poles of the system are inherentlyunstable.
  • Feedback control can be used to stabilize both the angle and position.
  • Other approaches involve oscillatingthe support up and down.